test20180919 递归问题

题意

定义
\[
f(n)=\left\{
\begin{array}{}
1 & n=1\\
f(n-f(f(n-1)))+1 & n>1
\end{array}
\right.
\]
\(g(n)\) 为满足\(f(m) = n\)的最大的\(m\)。
给出\(n\)，求\(g(n) \mod 998244353\)，和\(g(g(n)) \mod 998244353\)。

对100%的数据，\(n \leq 10^9\)

分析

代码

#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<ctime>
#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<complex>
#pragma GCC optimize ("O0")
using namespace std;
template<class T> inline T read(T&x)
{
    T data=0;
    int w=1;
    char ch=getchar();
    while(!isdigit(ch))
    {
        if(ch=='-')
            w=-1;
        ch=getchar();
    }
    while(isdigit(ch))
        data=10*data+ch-'0',ch=getchar();
    return x=data*w;
}
typedef long long ll;
const int INF=0x7fffffff;

const ll MAXN=1e6+7,lim=1e6,mod=998244353;
ll f[MAXN]={3,1,2,2};
ll ans1=5,ans2=11;
ll n;

int main()
{
//  freopen("recursion.in","r",stdin);
//  freopen("recursion.out","w",stdout);
    read(n);
    if(n==1)
    {
        puts("1 1");
        return 0;
    }
    if(n==2)
    {
        puts("3 5");
        return 0;
    }
    for(int i=3;i<=lim;++i)
    {
        for(int j=f[0]+1;j<=min(lim,f[0]+f[i]);++j)
            f[j]=i;

        ll up=min(f[i]+f[0],n),down=f[0]+1;
        if(down>n)
            break;

        (ans1 += (ll)(up - down + 1) * i)%=mod;

        if((up + down)&1)
        {
            (ans2 += (up + down) % mod * (up - down + 1)/2 % mod * i) %= mod;
        }

        else
        {
            (ans2 += (up + down)/2 % mod * (up - down + 1) % mod * i) %= mod;
        }

        f[0]+=f[i];
    }
    printf("%lld %lld\n",ans1,ans2);
//  fclose(stdin);
//  fclose(stdout);
    return 0;
}